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Amplifier

The DC operating point Q for a JFET amplifier is frequently chosen so that

$$I_D = \frac{I_{DSS}}{2}$$ (118)

i.e. midway in its operating range.

Now let's add an AC voltage source to the circuit of Figure 123, as in Figure 130 (this is not a practical amplifier circuit, it is used here for illustration only).

  

Figure 130: n-channel JFET as an amplifier.

We use the AC small signal model of Section 9.3.2, which we use to re-draw the circuit of Figure 130, see Figure 131.

  

Figure 131: AC equivalent circuit.

Now since i_g=0, we have v_g=v_gs=v_in. Then g_m v_gs = g_m v_in, and by KVL,

v_d = - i_d R_D = - g_m R_D v_in

and so the gain is given by

$$A = \frac{v_d}{v_{in}} = -g_m R_D .$$ (119)

This gain is negative, and depends on the transconductance g_m. The amplifier is an inverting amplifier.  

Example. Let R_D=1.2 k$\Omega$, V_DD=30 V, V_GG=-0.4 V, with the JFET parameters I_DSS=20 mA, V_p=-1.5 V. Then V_GS=-0.4 V, so

$$I_D = 20 \times 10^{-3} (1-\frac{-0.4}{-1.5})^2 = 10.76 \ mA$$

and

$$V_{DS} = 30 - (10.76 \times 10^{-3} )(1.2 \times 10^3) = 17.1 \ V$$

This defines the DC operating point Q.

The transconductance is

$$g_m = - \frac{2 \times 20 \times 10^{-3}}{-1.5}(1-\frac{-0.4}{-1.5}) = 19.55 \times 10^{-3} \ A/V$$

and so the amplifier gain is

$$A= - (19.55 \times 10^{-3})(1.2\times 10^3) = -23.47$$

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